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Posted by on Tuesday, October 13, 2009 at 5:06pm.

just wondering if someoen could help me with this limit..:

lim arctan[(x^2 - 4)/(3x^2-6x)]
x->2

  • calculus - , Tuesday, October 13, 2009 at 5:26pm

    My first step in doing limits is to sub in the approach value, this will give me Lim arctan (0/0)
    This just about guarantees that the algebraic expression will factor and the offending factor will cancel.
    Sure enough
    lim arctan[(x^2 - 4)/(3x^2-6x)] x->2
    = lim arctan[(x+2)(x-2)]/[3x(x-2)]
    =lim arctan[(x+2)]/[3x]
    = lim arctan(2/3)

    now use your calculator, set to radian mode, to find
    lim arctan[(x^2 - 4)/(3x^2-6x)] x->2
    = .588

  • calculus - , Tuesday, October 13, 2009 at 5:30pm

    thanks for the reply... the question asks me to evaluate the limit using continuity, how do i justify my answer with continuity?

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